如何使用if条件来使用css为以下代码标记分页活动页面？

Question

I am trying to put an if condition like <?php if($page===$x) {echo ' class="selected"';}?> to use CSS for marking the selected page in my pagination for the following code:

<?php
$result = $DB_con->query('SELECT COUNT(*) FROM tbl_users');
$totalRecords = $result->fetchColumn();
$totalPages = max(1, ceil($totalRecords / $recordsPerPage));

for ($i=1; $i<=$totalPages; $i++) {
echo "<span class='navigations_items_span'>";

//if($page===$i) echo "<b>Visiting Page</b>"; 
echo "<b>Page </b>";
echo "<a href='index3.php?page=".$i."' class='navigation_item selected_navigation_item'>. ".$i.".</a>";


echo "</span>";
};
?>

I wanted to place in the line : echo "<a href='index3.php?page=".$i."' class='navigation_item selected_navigation_item'>. ".$i.".</a>"; but getting error each time, how to place if here?

please help.

Answer 1

inside your for loop :

if($page == $i)
    $class = 'navigation_item selected_navigation_item';
else
    $class = '';

echo "<a href='index3.php?page=".$i."' class='".$class."'>. ".$i.".</a>";